Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it

II. (25 points) As shown in Figure 3, given the right triangle $\triangle ABC$ with the incircle $\odot I$ touching the two legs $AC$ and $BC$ at points $D$ and $E$ , respectively, and $AI$ , $BI$ intersecting line $DE$ at points $F$ and $G$ . Prove:
$AB^{2}=2 FG^{2} \text {. }$

Solution

As shown in Figure 7, connect $A G$ , $B F$ , $D I$ , and $E I$ .

Obviously, quadrilateral $CDIE$ is a square. Therefore,
$\begin{array}{l} \angle A D F = \angle A I B \\ = 135^{\circ}, \\ \angle F A D = \angle I A B . \end{array}$

Thus, $\triangle A D F \sim \triangle A I B$ .
Hence, $\frac{A D}{A I} = \frac{A F}{A B}$ .
Since $\angle D A I = \angle F A B$ , we have $\triangle A I D \sim \triangle A B F$ .
Therefore, $\angle A F B = \angle A D I = 90^{\circ}$ .
Similarly, $\angle A G B = 90^{\circ}$ .
Thus, points $A$ , $B$ , $F$ , and $G$ are concyclic, with $A B$ being the diameter of the circle, and
$\angle F B G = \angle A I B - \angle A F B = 45^{\circ}$ .
By the Law of Sines, we have
$F G = A B \sin \angle F B G \text{. }$

Therefore, $A B^{2} = 2 F G^{2}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.